列表作为函数参数 - 修改丢弃 [英] List as a function argument - modifications discarded

问题描述

我有以下code

def hidePasswords(L, password):
    for elem in L:
        if elem == password:
            elem = "*"*len(password)
    return L

print(hidePasswords(["test","test1","test8"],"test"))

它返回 ['测试'，'TEST1'，'test8'] 而不是 ['****'，'TEST1' test8'] 。当我改变我的功能

It returns ['test', 'test1', 'test8'] instead of ['****', 'test1', 'test8']. When I change my function to

def hidePasswords(L, password):
    temp = []
    for elem in L:
        if elem == password:
            elem = "*"*len(password)
        temp.append(elem)
    return temp

它工作正常。

为什么Python的以这样的方式表现？我完全理解在第二code会发生什么，但我不明白的第一个行为。

Why does Python behaves in such a way? I perfectly understand what happens in second code, but I don't understand behaviour of the first one.

推荐答案

在此code：

def hidePasswords(L, password):
    for elem in L:
        if elem == password:
            elem = "*"*len(password)
    return L

在循环中， ELEM 正在以→设置为每个值。然后重新分配 ELEM 是不同的东西。因此，它的已的在 L中的值，则是另一回事。这并不影响值→，因为 ELEM 不是一个参考。

In the loop, elem is being set to each value in L. Then you reassign elem to be something different. So it was the value in L, then it is something else. That doesn't affect the value in L, because elem isn't a reference.

您可以更简洁地写为：

def hidePasswords(L, password):
    return ['*'*len(password) if elem==password else elem for elem in L]

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